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Let $S$ be the set of all sequences $x=\{x_n\}$ of real numbers such that only a finite number of the $x_n$ are nonzero. Define $d(x,y)=\max|x_n-y_n|$. Is the space complete?

Completeness means that any Cauchy sequence converges to a point in the space. Suppose we have a sequence $\{\textbf{a}_i\}_{i=0}^\infty$, where $\textbf{a}_i=(a_{i1},a_{i2},\ldots)$. Since $\{\textbf{a}_i\}$ is Cauchy, for any fixed $k$, the real sequence $a_{1k},a_{2k},\ldots$ is also Cauchy, hence converges to a real number $b_i$. I'm not sure, however, whether the sequence $\{\textbf{a}_i\}$ converges to the sequence $\textbf{b}=(b_1,b_2,\ldots)$, or whether the sequence $\textbf{b}$ belongs to $S$.

Paul S.
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2 Answers2

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HINT: What if the first $n$ terms of $\mathbf{a}_n$ are the first $n$ terms of the harmonic series, the remaining terms all being $0$?

$$\begin{align*} \mathbf{a}_1&=\langle 1,0,0,0,\ldots\rangle\\ \mathbf{a}_2&=\left\langle 1,\frac12,0,0,0,\ldots\right\rangle\\ \mathbf{a}_3&=\left\langle 1,\frac12,\frac13,0,0,0,\ldots\right\rangle\\ &\;\vdots \end{align*}$$

Brian M. Scott
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  • Why exactly did you choose the harmonic series? Shouldn't any not-nearly constant null sequence work ? – Dominic Michaelis Aug 28 '13 at 20:42
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    @Dominic: Ease of description and because it’s familiar. Yes. – Brian M. Scott Aug 28 '13 at 20:43
  • @Brian since $l^{\infty}$ is complete and $S$ is a subspace,so $S$ is complete iff $S$ is closed, but since your example has a sequence that converges to a limit sequence with infinitely many terms as non-zero, hence the limit doesn't belong to $S$, therefore $S$ is not closed and hence not complete. – Upstart Jul 24 '19 at 20:06
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Here is a completely different approach than the one given by Brian.

Note that your space is isomorphic to $\Bbb R[x]$, and the metric is the supremum norm. The dimension of this space is $\aleph_0$, and therefore the space cannot be complete. Because then it is a Banach space, but if a Banach has a countable dimension, then it has a finite dimension.

Asaf Karagila
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